Question

if for x € (0,1/4), the derivative of tan^-1(6x√x/1-9x³) is √x.g(x) then g(x) is equal to ?

Answer 1

$let \: y = \tan^{ - 1} ( \frac{6x \sqrt{x} }{1 - 9 {x}^{3} } ) \:$

where x ∈ (0,1/4)

${ \tan }^{ - 1} ( \frac{2 \times (3 {x}^{ \frac{3}{2} }) }{1 - ( {3x}^{ \frac{3}{2} }) } ) = 2 { \tan }^{ - 1}(3 {x}^{ \frac{3}{2} })$

As 3x^3/2 ∈ (0,3/8)

$therefore \: \frac{dy}{dx} = 2 \times \frac{1}{1 + 9 {x}^{3} } \times 3 \times \frac{3}{2} \times {x}^{ \frac{1}{2} } \\ = \frac{9}{1 + {x}^{3} } \sqrt{3 } \\ therefore \: g(x) = \frac{9}{1 + {x}^{3} } \sqrt{3 } \:$